Imagine you have two shift-invariant measures $\mu, \nu$ in the Bernoulli space $\{0,1\}^{\mathbb{N}}$ with positive entropy and both are not the Bernoulli measure $(\frac{1}{2},\frac{1}{2})$. I know that:

$$ H( \mu \ast \nu) \geq \max(H( \mu ),H( \nu)) $$

Question. How can I prove this result with $>$ instead of $\geq$?

Thanks for your attention.

  • $\begingroup$ Only you to know, the group structure I'm considering here is the Z_{2}^{\N}, given two sequences x and y, the sequence x+y is the sequence: x+y = (x1+y1 mod2, x2+y2 mod2,...) $\endgroup$ – Bruno Brogni Uggioni Jun 10 '15 at 17:04
  • $\begingroup$ Do we have the inequality $H(\mu *\nu) \geq \max (H(\mu),H(\nu))$ for any positive measures in a compact group $G$ ? $\endgroup$ – BigBill Oct 18 '20 at 6:21

It is easy on some particular cases. Hope these simple cases could be generalized.

Suppose that $p\in[0,1]$ and call by $\mu^{(p)}$ the $(p,1-p)$-Bernoulli measure. Let $0\leq q,p\leq 1.$ Then $\mu^{(p)}*\mu^{(q)}=\mu^{(pq+(1-p)(1-q))}.$ In particular, if $q=0.5$ (or $p=0.5$) then $\mu^{(p)}*\mu^{(q)}=\mu^{(0.5)},$ and if $q=0,$ then $\mu^{(p)}*\mu^{(q)}=\mu^{(p)}$ (this is always true, but nice to check in this example). Using this observation, suppose that $0<q<p<1$ and $q,p\neq 0.5,$ then (an easy calculation gives that) $$H(\mu^{(p)}*\mu^{(q)})>\max\{H(\mu^{(p)}),H(\mu^{(q)})\}.$$

It is fun to consider $\epsilon>0$ small, and consider the function $$[0,1]\ni p\mapsto F(p):=H(\mu^{(p)}*\mu^{(\epsilon)}).$$ It has a unique maximum at $p=0.5$ with $F(0.5)=\log 2.$ It is monotone increasing on $[0,0.5],$ monotone decreasing on $[0.5,1]$ and $F(0)=F(1)=H(\mu^{(\epsilon)}).$

  • $\begingroup$ Dear @user39115, you are completely right, once that when we do the convolution of two Bernoulli measures we still have a Bernoulli measure. But, for instance, this does not work with Markov measures. I don't know, for instance, if the convolution of two Markov measures will have finite memory... but thanks for your answer $\endgroup$ – Bruno Brogni Uggioni Nov 12 '15 at 16:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.